Question

the doubling period of a bacterial population is 15 minutes. at time ( t = 100 ) minutes, the bacterial population was 80000. what was the initial population at time ( t = 0 )? find the size of the bacterial population after 5 hours.

Explanation:

Step1: Determine the general formula for population growth

The formula for exponential growth with a doubling period is \(P(t)=P_0\cdot2^{\frac{t}{d}}\), where \(P(t)\) is the population at time \(t\), \(P_0\) is the initial population, and \(d\) is the doubling period. Here \(d = 15\) minutes.

Step2: Find the initial population \(P_0\)

We know that when \(t = 100\) minutes, \(P(100)=80000\). Substitute into the formula:
\(80000=P_0\cdot2^{\frac{100}{15}}\)
\(80000=P_0\cdot2^{\frac{20}{3}}\)
\(P_0=\frac{80000}{2^{\frac{20}{3}}}\)
\(2^{\frac{20}{3}}=2^{6+\frac{2}{3}}=2^6\times2^{\frac{2}{3}} = 64\times\sqrt[3]{4}\approx64\times1.5874 = 101.5936\)
\(P_0=\frac{80000}{101.5936}\approx787\)

Step3: Convert 5 hours to minutes

Since 1 hour = 60 minutes, 5 hours \(t=5\times60 = 300\) minutes.

Step4: Find the population after 300 minutes

Use the formula \(P(t)=P_0\cdot2^{\frac{t}{d}}\), with \(P_0\approx787\), \(t = 300\), \(d = 15\)
\(P(300)=787\cdot2^{\frac{300}{15}}\)
\(P(300)=787\cdot2^{20}\)
\(2^{20}=1048576\)
\(P(300)=787\times1048576 = 825239312\)

Answer:

Initial population at \(t = 0\): \(787\)
Population after 5 hours: \(825239312\)